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anonymous
 3 years ago
A grain silo is shown below.
What is the volume of grain that could completely fill this silo rounded to the nearest whole number? Use 22/7 for pi.
19,008 ft3
19,461 ft3
6,336 ft3
453 ft3
anonymous
 3 years ago
A grain silo is shown below. What is the volume of grain that could completely fill this silo rounded to the nearest whole number? Use 22/7 for pi. 19,008 ft3 19,461 ft3 6,336 ft3 453 ft3

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sum together half of a sphere for the top part, with a cylinder for the bottom part. That's all there is to it :) Do you know the formulas for a cylinder and a sphere? h = 168 ft r = 6 ft

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large V_{sphere} = \frac{4 \pi r ^3}{3}\] \[\large V_{cylinder} = \pi r^2 h\] \[V_{total} = \frac{V_{sphere}}{2} + V_{cylinder}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you solve it now? ;D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no i'm sorry i'm completely lost :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0r = radius (both for the cylinder and sphere) , h = height of the cylinder Separate the curved, hemispherical top from the cylinder under it. You'll just add volume the two shapes together. What are you confused on specifically?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i'm just naturally confused on everything that has to do with maths you have to talk to me like a toddler and tell mi step by step

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you identify the radius in the drawing? And the height?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the radius for the entire drawing?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There's only one radius, yes. It works for both the sphere and cylinder, like I said.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes... dw:1342123496209:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And height is "h", which is?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.02/3*22/7*6*6*6 + 1/3*@22/7*6*6*168 = ............

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok, now scroll up and put r & h into those two equations for the Volume of a Sphere and Cylinder. Substitute, replacing r with "(6 ft)", and replacing h with "(168 ft)".

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1342123834485:dw find the volume of given figure

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@abhishekjha29 no, it's a cylinder under a hemisphere, not a cone.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ya tats ok.but i have a new figure wid a cone instead of culinder

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i use 3/4 3.14 times 6 to the 3rd power to find the volume of the sphere?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0??? @abhishekjha29 the volume of a cone = \(\large\frac{\pi r ^2 h}{3}\) , exactly a third of a cylinder's volume with the same dimensions for height and radius.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when i do it it tells me that the volume for the cylinder is 6330.24

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@rudyjanay I would leave the 3.1415... \(\pi\) out of it until the end, just keep it along like it's a unit like "feet". You can indeed do that above, but keep in mind it will be an estimate because: \(\pi \approx 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 062...\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Best to just leave it alone until you're all done with the formulas :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Kind of like how I did in the cylinder part above, if you see what I did

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When you're all done, multiply that big, nondecimal number and it'll give you your messy decimal answer.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok i'll try leaving the 3.14 out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0:) I think you've got it down now It's just plug & chug once you got the setup up done. The tricky part was recognizing it's half of a sphere (a hemisphere) plus a cylinder added together, and that the cylinder and the hemisphere have the same radius.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh i guess my phone was giving me the wrong answer because i don't have a calculator with me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the answer that i got is 19 461

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I just noticed I made a typo let me correct myself... \[\large V_{cylinder} = \pi r^2 h\] \[\large V_{cylinder} = \pi (6 \ ft)^2 (168 \ ft)\] \[\large V_{cylinder} = \pi (36 \ ft^2)(168 \ ft)\] \[\large V_{cylinder} = 6048\pi \ ft^3\] And back into the total... \[\large V_{total} = \frac{288\pi \ ft^3}{2} + (6048\pi ft^3)\] \[\large V_{total} = 144\pi \ ft^3 + 6048\pi ft^3\] \[\large V_{total} = 6192\pi \ ft^3 \approx \ 19452.74171... ft^3\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So yes, you're good. The answer choice you were given is actually an estimate, and not as exact as what I just did :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohhhh great thank you

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.06292\(\pi\) ft\(^3\) is the exact answer, and what I would be expect to do at work

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because if a construction firm wants to make an estimate to their liking, they can just pull out their pocket calculator or smartphone and see how many decimal places they care about for the project

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sometimes accuracy is critical and a matter of life & death, other times you don't have to be so particular

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thats why i don't get maths

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Example of where accuracy is a matter of life & death: ammunition factory. Even the smallest screw up could mean a detonation, a dud bullet in a critical field operation that means some cop or soldier gets killed, contracts being canceled etc. Another good example would be a factory where they make care engines. If they get the radius slightly wrong your engine won't work well or is more prone to breakdowns. Math can seem useless until you have something that makes it mean something :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Car engines* lol not "care"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you a math wiz or something

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Engineering student. I'm not mathematician, but I love the application of it to stuff that matters :3 Fire & explosives for example is a specialty of mine, and game design is my hobby (which also uses math quiet a bit because of physics stuff).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Math itself though, meh... I loose focus if it doesn't mean anything practical.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok i'm gonna check on my other question
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